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    In order to determine the depth of a well, a stone is dropped into the well and the time taken for the stone to drop is measured. It is found that the sound of the stone hitting the water arrives 5 seconds after the stone is dropped. If the speed of sound is 340ms-1. Find the depth of the well.

    I am really stuck on this question....

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    The answer is 107.5
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    Speed = distance/time ?
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    The depth is equal to the speed of sound times the time taken by the sound:

    340 m/s x 5 seconds
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    (Original post by Hilaryisme)
    The depth is equal to the speed of sound times the time taken by the sound:

    340 m/s x 5 seconds
    No.

    5s is the time for the stone to drop and the sound to reach the top again.

    Start with s=\frac{gt^2}{2} (from s=u+\frac{at^2}{2}) and t=5-\frac{s}{340}
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    You guys are really not smart. You can't do this without SUVAT. I'm in AS Level, so i'm not a master at this, but its the route you need to take.
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    Has anyone actually got the answer to the question?
    The answer is in the spoiler in the first post.
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    I've got the answer (if assuming the 'bottom' of the well is where the water is?)

    Why do you need help if you've got the answer?
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    (Original post by Squiidsquid)
    I've got the answer (if assuming the 'bottom' of the well is where the water is?)

    Why do you need help if you've got the answer?
    Likely he doesn't know the method, just has access to the answer.
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    Equate the distance the rock falls and the distance the sound travels back up. You should get an equation linking the times taken for both scenarios. From there you can work out a value for one of the times, and then evaluate the answer from that?
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    (Original post by gcse0)
    Has anyone actually got the answer to the question?
    The answer is in the spoiler in the first post.
    Yes. The equations that I gave lead to that answer:
    (Original post by RogerOxon)
    Start with s=\frac{gt^2}{2} (from s=u+\frac{at^2}{2}) and t=5-\frac{s}{340}
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    (Original post by RogerOxon)
    Yes. The equations that I gave lead to that answer:
    Could you explain how you obtain the equation, like the maths behind it?
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    (Original post by gcse0)
    Could you explain how you obtain the equation, like the maths behind it?
    and then it came to troy deeeennnnnnnnnnnnnnneyyyyyyyyyyy yyyyyyyyyyy
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    (Original post by gcse0)
    Could you explain how you obtain the equation, like the maths behind it?
    s=\frac{gt^2}{2}, \therefore t^2=\frac{2s}{g}

    \therefore (5-\frac{s}{340})^2=\frac{2s}{g}

    Simplify and solve the quadratic, taking the positive solution. You can confirm this by substituting the answer in the first post's spoiler into the above - it works.
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    (Original post by RogerOxon)
    s=\frac{gt^2}{2}, \therefore t^2=\frac{2s}{g}

    \therefore (5-\frac{s}{340})^2=\frac{2s}{g}

    Simplify and solve the quadratic, taking the positive solution. You can confirm this by substituting the answer in the first post's spoiler into the above - it works.
    Does g represent acceleration of the dropped object?
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    (Original post by gcse0)
    Does g represent acceleration of the dropped object?
    Yes
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